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3 years ago

How often is water added to the Earth system?

1 answer:

4 0

<span>Water is never added to earth system. Water forever remains in the water cycle on earth, so it goes from the ground, to the air, to the rain, to the sea, and round and round continuously. This cycle means that there does not need to be new water added to the earth, because it recycles any water that already exists of its own accord.</span>

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A 0.144-kg baseball is moving toward home plate with a speed of 43 m/s when

algol [13]

I would say 648858. bc yes

4 0

3 years ago

A busy chipmunk runs back and forth along a straight line of acorns that has been set out between his burrow and a nearby tree.

saveliy_v [14]

Answer: a = 1.32m/s2

Therefore, the average acceleration is 1.32m/s2

Explanation:

Acceleration is the rate of change in the velocity per time

a = change in velocity/time

a = ∆v/t

average acceleration a = (v2 -v1)/t. ....1

Given;

Final velocity v2 = 1.63m/s

Initial velocity v1 = -1.15ms

time taken t = 2.11s

Substituting into eqn 1

a = [1.63 - (-1.15)]/2.11

a = (1.63+1.15)/2.11

a = 2.78/2.11

a = 1.32m/s2

Therefore, the average acceleration is 1.32m/s2

6 0

3 years ago

The astronomer who discovered the dwarf planet Eris suggests there might be another object far beyond the Kupier belt. If this P

PolarNik [594]

These objects would be classified as extreme trans Neptunian object (ETNO).

Explanation:

ETNO’s are the objects lying beyond the planet Neptune and orbiting the Sun. They follow a highly eccentric path which is tilted. ETNO has been grouped into three major according to their respective perihelia.

Within this region (beyond Neptune’s orbit), a hypothetical planet has been discovered. It was discovered following its gravitational effect on the other objects of Kuiper Belt (region beyond the orbit of the Neptune- the last planet of our Solar system)

The Planet is assumed to be around 2 times the Earth’s size and around 10 times heavier than Earth.

4 0

2 years ago

A 2400W electric toaster is connected to a standard 120 V wall outlet. A family

jolli1 [7]

Answer:

1) energy = 15.6 kWh, 2) total_cost = $ 2.03

Explanation:

1) The energy dissipated is the product of the power and the time of use In a month it was used t = 6.5 h and the power of the toaster is

P = 2400 W = 2,400 kW

energy = P t

energy = 2,400 6.5

energy = 15.6 kWh

using rounding to a decimal

energy = 15.6 kWh

2) The cost of energy is unit_cost = $ 0.13 / kWh

so the total cost

total_cost = energy unit_cost

total_cost = 15.6 0.13

total_cost = $ 2.028

rounding to two decimal places

total_cost = $ 2.03

5 0

2 years ago

Lasers are now used in eye surgery. Given the wavelength of a certain laser is 514 nm and the power of the laser is 1.1 W, how m

Leno4ka [110]

Answer: $1.593*10^{17} photons$ released if the laser is used 0.056 s during the surgery

Explanation:

First, you have to calculate the energy of each photon according to Einstein's theoty, given by:

$E =\frac{hc}{\lambda}$

Where $\lambda$ is the wavelength, $h$ is the Planck's constant and $h$ is the speed of light

$h = 6.626*10^{-34} \frac{m^{2} kg }{s}$ -> Planck's constant

$c = 3*10^{8} \frac{m}{s}$ -> Speed of light

So, replacing in the equation:

$E =\frac{ 6.626*10^{-34} \frac{m^{2} kg }{s}*3*10^{8} \frac{m}{s}}{514*10^-9 m}$

Then, the energy of each released photon by the laser is:

$E = 3.867*10^{-19} \frac{J}{photons}$

After, you do the inverse of the energy per phothon and as a result, you will have the number of photons in a Joule of energy:

$\frac{1}{3.867*10^{-19}} = 2.586*10^{18} \frac{photons}{J}$

The power of the laser is 1.1 W, or 1.1 J/s, that means that you can calculate how many photons the laser realease every second:

$2.586*10^{18}\frac{photons}{J} * 1.1 \frac{J}{s} = 2.844*10^{18} \frac{photons}{s}$

And by doing a simple rule of three, if $2.844*10^{18} photons$ are released every second, then in 0.056 s:

$0.056 s*2.844*10^{18} \frac{photons}{s} = 1.593*10^{17} photons$ are released during the surgery

5 0

3 years ago